Cambridge Maths Academy

16. Integration of powers of the sine function (Wallis integral) 본문

수학 모음 (Maths collection)/Technical A - Exploring ideas

16. Integration of powers of the sine function (Wallis integral)

Cambridge Maths Academy 2021. 12. 9. 08:29
반응형

수학 모음 (Maths collection) 전체보기

 

Question. Let In=sinn(ax)dx

(a) Show that In=1ansinn1(ax)cos(ax)+n1nIn2

 

Let Jn=π20sinnxdx

(b) Hence, or otherwise, show that Jn=n1nJn2 and hence that (i)n=2m(even):J2m=π20sin2mxdx=(2m)!22m(m!)2π2(ii)n=2m+1(odd):J2m+1=π20sin2m+1xdx=22m(m!)2(2m+1)!

 

Comment. We can generalise this result to: K2m=π2a0sin2m(ax)dx=(2m)!22m(m!)2π2aK2m+1=π2a0sin2m+1(ax)dx=22m(m!)2(2m+1)!1a

 

(Relevant: Edexcel 2017 Specification, FP2 Ch6 Reduction formulae)

 

 

Solution. (a) By integration by parts, u=sinn1(ax)v=sin(ax)u=a(n1)sinn2(ax)cos(ax)v=1acos(ax) we find In=sinn(ax)dx=1asinn1(ax)cos(ax)+(n1)sinn2(ax)cos2(ax)dx=1asinn1(ax)cos(ax)+(n1)sinn2(ax)(1sin2(ax))dx=1asinn1(ax)cos(ax)+(n1)sinn2(ax)dxIn2(n1)sinn(ax)dxIn=1asinn1(ax)cos(ax)+(n1)In2(n1)In[1+(n1)]In=1asinn1(ax)cos(ax)+(n1)In2nIn=1asinn1(ax)cos(ax)+(n1)In2In=1ansinn1(ax)cos(ax)+n1nIn2

 


 

(b) From (a) with a=1, Jn=π20sinnxdx=[1nsinn1xcosx]π20=0+n1nJn2=n1nJn2

 

For the next part, we may approach it in three different ways.

 

Method 1.

더보기

We note the following results: J0=π20dx=π2J1=π20sinxdx=[cosx]π20=cos(π2)+cos(0)=1J2=π20sin2xdx=π20(1cos2x2)dx=[12x14sin2x]π20=π4(=12J0) (i) n=2m (even) J2m=2m12mI2m2=2m12m2m32m2J2m4=2m12m2m32m22m52m4J2m6=2m12m2m32m22m52m43412J0π2=(2m1)(2m3)(2m5)312m(2m2)(2m4)42π2=(2m)(2m1)(2m2)(2m3)(2m4)(2m5)321[2m(2m2)(2m4)42]2π2=(2m)![2mm(m1)(m2)21]2π2=(2m)!22m(m!)2π2 (ii) n=2m+1 (odd) J2m+1=2m2m+1J2m1=2m2m+12m22m1J2m3=2m2m+12m22m12m42m3J2m5=2m2m+12m22m12m42m321J1=1=2m(2m2)(2m4)42(2m+1)(2m1)(2m1)31=[2m(2m2)(2m4)42]2(2m+1)(2m)(2m1)(2m2)(2m3)(2m4)321=[2mm(m1)(m2)21]2(2m+1)!=22m(m!)2(2m+1)!

 

Aside. As an exemplary application, we evaluate the first few integrals: J0=π20dx=π2J1=π20sinxdx=1J2=π20sin2xdx=π4J3=π20sin3xdx=23J4=π20sin4xdx=3π16J5=π20sin5xdx=815J6=π20sin6xdx=5π32J7=π20sin7xdx=1635J8=π20sin8xdx=35π256J9=π20sin9xdx=128315J10=π20sin10xdx=63π512J11=π20sin11xdx=256693

 


 

Method 2. Exponential form and binomial theorem:

더보기

We can also use the exponential definition of the sine function, sinx=eixeix2i (i) n=2m (even) \begin{align} J_{2m} &= \int_0^{ \frac{ \pi }{ 2 } } \sin^{2m} x \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \left( \frac{ \textrm{e}^{ix} - \textrm{e}^{-ix} }{ 2i } \right)^{2m} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m}} \left( \textrm{e}^{ix} - \textrm{e}^{-ix} \right)^{2m} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m}} \left[ \sum_{r=0}^{2m} \binom{2m}{r} \textrm{e}^{i(2m-r)x} (-1)^r \textrm{e}^{-irx} \right] \, \textrm{d}x \qquad \textrm{by binomial theorem} \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} (-1)^r \underbrace{ \left( \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{2i(m-r)x} \, \textrm{d}x \right) }_{ I_r } \\ \end{align}

 

Here, we note that, for r \ne m, \begin{align} I_r &= \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{2i(m-r)x} \, \textrm{d}x \\ &= \left[ \frac1{2i(m-r)} \textrm{e}^{2i(m-r)x} \right]_0^{ \frac{ \pi }{ 2 } } \\ &= \frac{1}{2i(m-r)} \left( \textrm{e}^{i(m-r)\pi} - 1 \right) \\ \end{align} which gives (ignoring the subtlety with r=m for the next few lines as we're just trying to see the pattern for now) \begin{align} \Rightarrow \qquad J_{2m} &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} (-1)^r I_r \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} \frac{(-1)^r}{2i(m-r)} \left( \textrm{e}^{i(m-r)\pi} - 1 \right) \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} \frac{(-1)^r}{2i(m-r)} \Big( \cos[(m-r)\pi] - 1 \Big) \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} \frac{(-1)^r(-1)}{i(m-r)} \sin^2 \left( \frac{(m-r)\pi}{2} \right) \end{align} We observe a symmetry in the terms with r=s and r=2m-s, \begin{align} r&=s \;:& a_s&=\binom{2m}{s} \frac{(-1)^s(-1)}{i(m-s)} \sin^2 \left( \frac{(m-s)\pi}{2} \right) \\ \\ r&=2m-s \;:& a_{2m-s}&=\binom{2m}{2m-s} \frac{(-1)^{2m-s}(-1)}{i(s-m)} \sin^2 \left( \frac{(s-m)\pi}{2} \right) \end{align} and, since \binom{2m}{s}=\binom{2m}{2m-s}, we find \begin{align} a_s = - a_{2m-s} \end{align} This motivates to separate the sum into three parts, \begin{align} \sum_{r=0}^{2m} = \sum_{r=0}^{m-1} + \sum_{r=m}^{m} + \sum_{r=m+1}^{2m} \end{align} and in fact we should apply this separation before we perform the integral of I_r and include the case r=m as well: \begin{align} \Rightarrow \qquad J_{2m} &= \frac1{(2i)^{2m}} \left( \sum_{r=0}^{m-1} + \sum_{r=m}^{m} + \sum_{r=m+1}^{2m} \right) \binom{2m}{r} (-1)^r I_r \\ &= \frac{(-1)^m}{2^{2m}} \left[ \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r I_r + \binom{2m}{m} (-1)^m I_m + \sum_{r=m+1}^{2m} \binom{2m}{r} (-1)^r I_r \right] \end{align} For the last summation, let k = 2m-r \begin{align} \Rightarrow \qquad J_{2m} &= \frac{(-1)^m}{2^{2m}} \left[ \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r I_r + \binom{2m}{m} (-1)^m I_m + \sum_{k=0}^{m-1} \binom{2m}{2m-k} (-1)^{2m-k} I_{2m-k} \right] \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r \Big( I_r + I_{2m-r} \Big) + \frac{(-1)^m}{2^{2m}} \binom{2m}{m} (-1)^m I_m \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r \left( \frac{\textrm{e}^{i(m-r)\pi} - 1}{2i(m-r)} + \frac{\textrm{e}^{-i(m-r)\pi} - 1}{-2i(m-r)} \right) + \frac{1}{2^{2m}} \binom{2m}{m} \frac{\pi}{2} \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r \underbrace{ \left( \frac{ \cos[(m-r)\pi] - 1}{2i(m-r)} - \frac{\cos[(m-r)\pi] - 1}{2i(m-r)} \right) }_{ =0 } + \frac{1}{2^{2m}} \binom{2m}{m} \frac{\pi}{2} \\ &= \frac{1}{2^{2m}} \binom{2m}{m} \frac{\pi}{2} \\ &= \frac{ (2m)! }{ 2^{2m} (m!)^2 } \frac{\pi}{2} \qquad \square \end{align}

 

(ii) n=2m+1 (odd) \begin{align} J_{2m+1} &= \int_0^{ \frac{ \pi }{ 2 } } \sin^{2m+1} x \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \left( \frac{ \textrm{e}^{ix} - \textrm{e}^{-ix} }{ 2i } \right)^{2m+1} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m+1}} \left( \textrm{e}^{ix} - \textrm{e}^{-ix} \right)^{2m+1} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m+1}} \left[ \sum_{r=0}^{2m+1} \binom{2m+1}{r} \textrm{e}^{i(2m+1-r)x} (-1)^r \textrm{e}^{-irx} \right] \, \textrm{d}x \qquad \textrm{by binomial theorem} \\ &= \frac1{(2i)^{2m+1}} \sum_{r=0}^{2m+1} \binom{2m+1}{r} (-1)^r \underbrace{ \left( \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{i(2m+1-2r)x} \, \textrm{d}x \right) }_{ I_r } \\ \end{align} We note that, \begin{align} I_r &= \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{i(2m+1-2r)x} \, \textrm{d}x \\ &= \left[ \frac1{i(2m+1-2r)} \textrm{e}^{i(2m+1-2r)x} \right]_0^{ \frac{ \pi }{ 2 } } \\ &= \frac1{i(2m+1-2r)} \left( \textrm{e}^{i(m-r)\pi + \frac{i\pi}{2} } - 1 \right) \\ &= \frac{ i \cos [(m-r)\pi] - 1 }{ i(2m+1-2r) } \\ \\ I_{2m+1-r} &= \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{-i(2m+1-2r)x} \, \textrm{d}x \\ &= \left[ - \frac1{i(2m+1-2r)} \textrm{e}^{i(2m+1-2r)x} \right]_0^{ \frac{ \pi }{ 2 } } \\ &= - \frac1{i(2m+1-2r)} \left( \textrm{e}^{-i(m-r)\pi - \frac{i\pi}{2} } - 1 \right) \\ &= - \frac{ - i \cos [(m-r)\pi] - 1 }{ i(2m+1-2r) } \\ &= \frac{ i \cos [(m-r)\pi] + 1 }{ i(2m+1-2r) } \\ \end{align} As with the even case above, we separate the summation into two parts, \begin{align} \sum_{r=0}^{2m+1} = \sum_{r=0}^{m} + \sum_{r=m+1}^{2m+1} \end{align} \begin{align} \Rightarrow \qquad J_{2m+1} &= \frac1{(2i)^{2m+1}} \left( \sum_{r=0}^{m} + \sum_{r=m+1}^{2m+1} \right) \binom{2m+1}{r} (-1)^r I_r \\ \end{align} For the second summation, let k = 2m+1-r and it gives \begin{align} \Rightarrow \qquad J_{2m+1} &= \frac{1}{(2i)^{2m+1}} \left[ \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r I_r + \sum_{k=0}^{m} \binom{2m+1}{2m+1-k} (-1)^{2m+1-k} I_{2m+1-k} \right] \\ &= \frac{1}{(2i)^{2m+1}} \left[ \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r I_r + \sum_{k=0}^{m} \binom{2m+1}{k} (-1)^{k-1} I_{2m+1-k} \right] \\ &= \frac{1}{(2i)^{2m+1}} \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r \Big( I_r - I_{2m+1-r} \Big) \\ &= \frac{1}{(2i)^{2m+1}} \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r \left( \frac{ i \cos [(m-r)\pi] - 1 }{ i(2m+1-2r) } - \frac{ i \cos [(m-r)\pi] + 1 }{ i(2m+1-2r) } \right) \\ &= \frac{1}{(2i)^{2m+1}} \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^{r+1} \left( \frac{ 2 }{ i(2m+1-2r) } \right) \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m} \binom{2m+1}{r} \frac{ (-1)^{r} }{ (2m+1-2r) } \qquad \textrm{(Dirichlet beta function?)}\\ &= \frac{(-1)^m}{2^{2m}} \left( \frac{ 2^{2m+1} \Gamma \left( \frac12 - m \right) \Gamma(m+1) }{ \sqrt{\pi} (2m+1) } \right) \\ &= \frac{ 2(-1)^m m! }{ \sqrt{\pi} (2m+1) } \Gamma \left( \frac12 - m \right) \\ &= \frac{ 2(-1)^m m! }{ \sqrt{\pi} (2m+1) } \left[ (-1)^m \frac{ 2^{2m} m! }{ (2m)! } \sqrt{ \pi } \right] \\ &= \frac{ 2^{2m+1} (m!)^2 }{ (2m+1)! } \qquad \square \end{align} since \begin{align} \Gamma \left( -m + \frac12 \right) &= \frac{ \Gamma \left( -m +1 + \frac12 \right) }{ \left( -m + \frac12 \right) } \\ &= \frac{ \Gamma \left( -m + m + \frac12 \right) }{ \left( -m + \frac12 \right) \left( -m + 1 + \frac12 \right) \left( -m + 2 + \frac12 \right) \cdots \left( -m + (m - 1) + \frac12 \right) } \\ &= \frac{ \Gamma \left( \frac12 \right) }{ (-1)^m \frac{ (2m-1)(2m-3)(2m-5) \cdots 3 \cdot 1 }{ 2^m } } \\ &= \frac{ \sqrt{\pi} }{ (-1)^m \frac{ (2m)(2m-1)(2m-2)(2m-3)(2m-4) \cdots 1 }{ 2^m (2m)(2m-2)(2m-4) \cdots 2 } } \\ &= \frac{ \sqrt{\pi} }{ (-1)^m \frac{ (2m)! }{ 2^{2m} m! } } \\ &= (-1)^m \frac{ 2^{2m} m! }{ (2m)! } \sqrt{ \pi } \end{align}

 


 

Method 3. Integration by substitution and the Euler beta and gamma functions:

더보기

Use the substitution \begin{align} u &= \sin x \\ \textrm{d}u &= \cos x \, \textrm{d}x = \sqrt{ 1 - \sin^2x }\, \textrm{d}x = \sqrt{ 1 - u^2 }\, \textrm{d}x \end{align} and the integral reads \begin{align} \Rightarrow \qquad J_n &= \int_0^{\frac{\pi}{2}} \sin^nx \, \textrm{d}x = \int_0^1 u^n \left( 1 - u^2 \right)^{-\frac12} \, \textrm{d}u \end{align} We make the substitute: \begin{align} t &= u^2 \\ \textrm{d}t &= 2u \, \textrm{d}u = 2t^{\frac12} \, \textrm{d}u \end{align} which gives \begin{align} \Rightarrow \qquad J_n &= \int_0^1 t^{ \frac{n}{2} } \left( 1 - t \right)^{-\frac12} \, \frac{ \textrm{d}t }{ 2t^{\frac12} } \\ &= \frac12 \int_0^1 t^{ \frac{n}{2} - \frac12 } \left( 1 - t \right)^{-\frac12} \, \textrm{d}t \\ &= \frac12 B \left( \frac{n+1}{2}, \frac12 \right) \\ &= \frac12 \frac{ \Gamma\left( \frac{n}{2} + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( \frac{n}{2}+1 \right) } \end{align} where the Euler beta and gamma functions are defined by \begin{align} B(p,q) &= \int_0^1 t^{p-1} (1-t)^{q-1} \, \textrm{d}t \\ \Gamma(p) &= \int_0^\infty t^{p-1}\textrm{e}^{-t} \, \textrm{d}t \end{align} and \begin{align} B(p,q) = \frac{ \Gamma(p) \Gamma(q) }{ \Gamma(p+q) } \qquad \textrm{and} \qquad \Gamma(p+1) = p \Gamma(p) \end{align} Noting that \Gamma \left( \frac12 \right) = \sqrt{\pi} (see below), we find, for m \in \mathbb N_0, \begin{align} \Gamma \left( m + \frac12 \right) &= \left( m - \frac12 \right) \Gamma \left( m - \frac12 \right) \\ &= \underbrace{ \left( m - \frac12 \right) \left( m - \frac32 \right) \cdots \left( \frac12 \right) }_{n\;\textrm{terms}} \Gamma \left( \frac12 \right) \\ &= \frac{ (2m-1) (2m-3) \cdots 3 \cdot 1 }{ 2^m } \Gamma \left( \frac12 \right) \\ &= \frac{ (2m) (2m-1) (2m-2) (2m-3) (2m-4) \cdots 3 \cdot 2 \cdot 1 }{ 2^m (2m) (2m-2) (2m-4) \cdots 4 \cdot 2 } \underbrace{ \Gamma \left( \frac12 \right) }_{=\sqrt{\pi}} \\ &= \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi} \end{align} where we have used \begin{align} \left[ \Gamma \left( \frac12 \right) \right]^2 &= B\left( \frac12, \frac12 \right) \underbrace{ \Gamma(1) }_{0!=1} \\ &= \int_0^1 t^{-\frac12} (1-t)^{-\frac12} \, \textrm{d}t \\ &= \int_0^1 \frac1{ \sqrt{ t(1-t)} } \, \textrm{d}t \\ &= \int_0^1 \frac1{ \sqrt{ \frac14 - \left( t - \frac12 \right)^2 }} \, \textrm{d}t \end{align} and by the substitution \begin{align} t-\frac12 &= \frac12\sin v \\ \textrm{d}t &= \frac12\cos v\,\textrm{d}v \\ \end{align} which gives \begin{align} &\Rightarrow& \left[ \Gamma \left( \frac12 \right) \right]^2 &= \int_0^{\frac{\pi}{2}} \frac1{ \sqrt{ \frac14 \left( 1 - \sin^2v \right) }} \, \frac12\cos v\,\textrm{d}v = \pi \\ &\Rightarrow& \Gamma \left( \frac12 \right) &= \sqrt{ \pi } \end{align}

 

(i) n=2m (even) \begin{align} J_{2m}&= \frac12 \frac{ \Gamma\left( \frac{2m}{2} + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( \frac{2m}{2}+1 \right) } \\ &= \frac12 \frac{ \Gamma\left( m + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( m+1 \right) } \\ &= \frac12 \frac{ \left( \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi} \right) \sqrt{\pi} }{ m! } \\ &= \frac{ (2m)! }{ 2^{2m} (m!)^2 } \frac{\pi}{2} \qquad \square \end{align}

 

(ii) n=2m+1 (odd) \begin{align} J_{2m+1}&= \frac12 \frac{ \Gamma\left( \frac{2m+1}{2} + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( \frac{2m+1}{2}+1 \right) } \\ &= \frac12 \frac{ \Gamma\left( m + 1 \right) \Gamma\left( \frac12 \right) }{ \Gamma \left( m + \frac32 \right) } \\ &= \frac12 \frac{ m! \sqrt{\pi} }{ \left( m + \frac12 \right) \Gamma \left( m + \frac12 \right) } \\ &= \frac12 \frac{ m! \sqrt{\pi} }{ \left( m + \frac12 \right) \left[ \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi} \right] } \\ &= \frac{ 2^{2m+1} (m!)^2 }{ (2m+1)! } \qquad \square \end{align}

 


 

A summary:

 

\begin{align} \int_0^{\frac{\pi}{2}}\sin^{2m}x\,\textrm{d}x=\frac{(2m)!}{2^{2m}(m!)^2}\frac{\pi}{2} \\ \\ \int_0^{\frac{\pi}{2}}\sin^{2m+1}x\,\textrm{d}x=\frac{2^{2m}(m!)^2}{(2m+1)!} \\ \\ \int_0^{\frac{\pi}{2}}\cos^{2m}x\,\textrm{d}x=\frac{(2m)!}{2^{2m}(m!)^2}\frac{\pi}{2} \\ \\ \int_0^{\frac{\pi}{2}}\cos^{2m+1}x\,\textrm{d}x=\frac{2^{2m}(m!)^2}{(2m+1)!} \end{align}

and \begin{align} \Gamma \left( \frac12 \right) &= \sqrt{ \pi } \\ \Gamma \left( m + \frac12 \right) &= \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi}, \quad m \in \mathbb Z_0^+ \\ \Gamma \left( -m + \frac12 \right) &= (-1)^m \frac{ 2^{2m} m! }{ (2m)! } \sqrt{ \pi }, \quad m \in \mathbb Z_0^+ \end{align}

반응형
Comments